Implement an iterator over a binary search tree (BST). Your iterator will be initialized with the root node of a BST.

Callingnext()will return the next smallest number in the BST.

Note:next()andhasNext()should run in average O(1) time and uses O(h) memory, wherehis the height of the tree.

Solution: BST用inorder traversal。实际就是把stack实现inorder traversal拆开在几个部分

    Stack<TreeNode> stack = new Stack<>();
    TreeNode curt;

    public BSTIterator(TreeNode root) {
        curt = root;
        while(curt != null){
                stack.push(curt);
                curt = curt.left;
        }
    }

    /** @return whether we have a next smallest number */
    public boolean hasNext() {
        return !stack.empty();
    }

    /** @return the next smallest number */
    public int next() {
        curt = stack.pop();
        int val = curt.val;
        curt = curt.right;
        while(curt != null){
                stack.push(curt);
                curt = curt.left;
        }
        return val;
    }